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8. (a) Define (i) acid, (ii) base, according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 8 - 2005

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8. (a) Define (i) acid, (ii) base, according to the Brønsted-Lowry theory. (b) Identify one species acting as an acid, and also identify its conjugate base, in the ... show full transcript

Worked Solution & Example Answer:8. (a) Define (i) acid, (ii) base, according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 8 - 2005

Step 1

Define (i) acid, (ii) base, according to the Brønsted-Lowry theory.

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Answer

According to the Brønsted-Lowry theory:

(i) An acid is defined as a substance that can donate a proton (H<sup>+</sup> ion) to another substance.

(ii) A base is defined as a substance that can accept a proton (H<sup>+</sup> ion) from another substance.

Step 2

Identify one species acting as an acid, and also identify its conjugate base, in the following system.

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Answer

In the provided equilibrium:

H<sup>+</sup> + CT ⇌ HCl + HF,

the species acting as an acid is HCl (Hydrochloric acid) because it donates a proton to form HF (Hydrofluoric acid). The conjugate base of HCl is Cl<sup>-</sup>.

Step 3

Calculate the pH of a 0.002 M solution of methanoic acid (HCOOH). The value of K<sub>a</sub> for methanoic acid is 1.8 × 10<sup>-4</sup>.

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To calculate the pH, we can start with the dissociation of methanoic acid:

HCOOH ⇌ H<sup>+</sup> + HCOO<sup>-</sup>.

Let

  • [H<sup>+</sup>] = x,
  • [HCOO<sup>-</sup>] = x,
  • [HCOOH] = 0.002 - x (approximately 0.002 since x is small).

The expression for K<sub>a</sub> is:

Ka=[H+][HCOO][HCOOH]K_a = \frac{[H^+][HCOO^-]}{[HCOOH]}

Substituting the values gives:

1.8×104=x20.0021.8 \times 10^{-4} = \frac{x^2}{0.002}

Solving for x gives:

x2=1.8×104×0.002=3.6×107x^2 = 1.8 \times 10^{-4} \times 0.002 = 3.6 \times 10^{-7} x=3.6×1070.0006x = \sqrt{3.6 \times 10^{-7}} \approx 0.0006

The concentration of hydrogen ions is [H<sup>+</sup>] = 0.0006 M.

Now calculate the pH:

pH=log[H+]=log(0.0006)3.22pH = -\log[H^+] = -\log(0.0006) \approx 3.22

Thus, the pH of the solution is approximately 3.22.

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